Your equations seem to indicate that the 70 degrees is with the FLOOR, not the Wall, which makes more sense)
First, force up = force down
so force up at floor = 160
Call friction force at floor Fw
It is equal and opposite to horizontal force at top
Now moments about ladder at floor sum to zero
160 cos 70 (L/2)= Fw sin 70(L)
80 = Fw sin 70/cos 70
BUT sin/cos = tan
so Fw = 80/tan 70 = 29.1 Newtons
A uniform 160 N ladder rests against a perfectly smooth wall, making a 70 degree angle with the wall (ladder is 7 m long). (a) find the normal forces that the wall and the floor exert on the ladder. B) what is the friction force on the ladder at the floor?
I tried to solve the problem and by making a sketch and go from there...
attempt: m1g(mass x gravity of ladder)cos(70)(7/2 is the lenght of the ladder divided by 2 because is in the center of the ladder that is exerted)= FW(force of the wall)sin(70)(7)...
160cos(70)(7/2)=FWsin(70)(7)
FW=[160cos(70)(7/2)]/[sin(70)(7)
from here on i do not know why my teacher told me to conver it to tan and he cross the lenghts but i do understand the reason for that, the only problem is the tan...
IT WILL BE VERY APPRECIATED IF SOMEBODY CAN EXPLAIN THE REASON FOR THE TAN PART
1 answer
1 answer